Conference rusure::math

     <<< Note 1512.0 by 3D::CORKUM "We'll be right there Miss Fletcher" >>>
                     -< Wronskian ( Bessel and Legendre ) >-

>> I have a Bessel of order zero:

>>	xy'' + y' + xy = 0  ( homogeneous )   y' = dy/dx       y'' = d2y/dx2

>> The problem states:

>> 1) Show that the Wronskian of any 2 solutions is W (x) = C / x

>>  (where C is a constant dependent on the particular choice of solutions.)

>> 2) What is the Wronskian of 2 solutions of the Legendre equation ?

>> 3)  "   "   "    "       "  "    "       "  a Bessel equation of order m ?

>> attempt at the answer #1)

>>	first determined roots:   ( double root at s = 0 )

>>	solutions:   y1 = const    
>>		     y2 = const * Log x  ( by variation of constants )

>>	my Wronskian = y1**2 / x    ( where C = y1**2 ?? )

>>	so can I just state that the constant is y1**2, for y1 is a constant ?

No, because your y1 and y2 don't satisfy Bessel's equation.

I'm not sure if this will satisfy your teacher, but here goes:

(Formulae and notation from M. E. Abramowitz and L. E. Stegun: Handbook of
mathematical functions (published by National Bureau of Standards (or whatever
they are called at the moment)).)

A basis for the solution space to Bessel's equation of order m is {Jm(z),Ym(z)}.
The Wronskian, by definition, is

	Wm(z) = Jm(z) * Ym'(z) - Jm'(z) * Ym(z)

Using 9.1.32 and 9.1.34 (with a == b == z), we get

	Pm(a,b) = Jm(a) * Ym(b) - Jm(b) * Ym(a)		(== 0 if a == b)
	Qm(a,b) = Jm(a) * Ym'(b) - Jm'(b) * Ym(a)
	Rm(a,b) = Jm'(a) * Ym(b) - Jm(b) * Ym'(a)
	Sm(a,b) irrelevant

	Pm(a,b) * Sm(a,b) - Qm(a,b) * Rm(a,b) = 4 / (pi**2 * a * b)

Now, if a == b == z, the above simplifies to

	Rm(z,z) ** 2 = 4 / (pi**2 * z * z)

Taking square roots (and checking the sign), we get the answer to questions (1)
and (3) as

	Wm(z) = 2 / (pi * z)

I suspect that the answer to question (2) is also to be found in Abramowitz and
Stegun.

    Peter,
    
    	Thank You for the answer.... I looked in over...
    	and it looks a lot better than my submitted attempt below...
    
        If anyone cares to comment on my attempt of the solution below, I'd
    	be glad to read them....
    		
    	I ended trying a different angle on solving the problem.  I found
    	a relationship between p(x) and the Wronskian....in my text and
    	proceeded blindly down a path at attempting to solve the problem.
    
    	
    	W = W(0) * exp ** integral of p(x) 
    
    	y'' +y'/x +y = 0 ( original bessel first kind, order 0 )
    
    	thus p(x) = 1/x
    
    	integrating yields ==> ln |x|
    
    	e to the power ln |x| = x  ( evaluated at X2 and X1 )
    
    	W = W(0) * integral of p(x) = W(0) * |x2|/|x1|
    
    	I made a big jump and let |x2| = 1 as P(x) is independent of path
    			  and let |x1| = some variable X
    
    	and came up with
    				W(0) * constant
    				---------------
    					X
    
    	I probably have gapping holes in my reasoning, but It was my
    attempt. I might have to pick up this reference you mentioned...to
    browse on through it...
    
    currently it seems like I'm acquiring a few volumes to get me through
    this course..... I've got Shaums, REA, kaplan, knoop, Kreysig, Butkov,
    and a bunch more reimann and conformal mapping texts....most are
    borrowed on on loan from a library....Probably 14 in all....
    

     <<< Note 1512.2 by 3D::CORKUM "We'll be right there Miss Fletcher" >>>

>>    Peter,
    
>>    	Thank You for the answer.

You're welcome.
    
>>	If anyone cares to comment on my attempt of the solution below, I'd
>>	be glad to read them....

>>	I ended trying a different angle on solving the problem.  I found
>>	a relationship between p(x) and the Wronskian....in my text and
>>	proceeded blindly down a path at attempting to solve the problem.
    
>>	W = W(0) * exp ** integral of p(x) 

**********
    
>>	y'' +y'/x +y = 0 ( original bessel first kind, order 0 )
    
>>	thus p(x) = 1/x
    
>>	integrating yields ==> ln |x|
    
>>	e to the power ln |x| = x  ( evaluated at X2 and X1 )
    
>>	W = W(0) * integral of p(x) = W(0) * |x2|/|x1|
    
>>	I made a big jump and let |x2| = 1 as P(x) is independent of path
>>			  and let |x1| = some variable X
    
>>	and came up with
>>				W(0) * constant
>>				---------------
    					X
    
>>	I probably have gaping holes in my reasoning, but it was my
>>    attempt. I might have to pick up this reference you mentioned...to
>>    browse on through it...

Your reasoning can be patched up:

Let I(x1, x2, f(...), dt) be the integral from x1 to x2 of the function f(...)
with respect to t.  I assume that the starred relation (W = ...) above should
be

    W(x) = W(0) * exp(I(x0, x, p(t), dt))

Let's assume x>0 and x0>0 � to get rid of the absolute sign.  Then we get

    W(x) = W(0) * x0 / x

which is of the desired form, since W(0) and x0 are constants.  (A detail:
because ln(x) has an essential singularity at the origin, we can't set x0 to
0. Interesting question: what happens if x0-->0+?)

>>    currently it seems like I'm acquiring a few volumes to get me through
>>    this course..... I've got Shaums, REA, kaplan, knoop, Kreysig, Butkov,
>>    and a bunch more reimann and conformal mapping texts....most are
>>    borrowed on on loan from a library....Probably 14 in all....

Expect to get a few more.  For real life, the reference books (such as
Abramowitz and Stegun) are *far* more useful than the text books, because
you'll have mastered the concepts (as provided by the text books), but will
need a convenient way of finding sundry (more or less) well-known results.

=================================

� note that, because of analytic continuation, we can restrict ourselves to a
suitable part of the real line and still be sure that the results we arrive at
will be valid everywhere (except, of course for singularities).

    Peter,
    
    	Thanks for the clarification. You were correct in your assumptions.
    
    
    >which is of the desired form, since W(0) and x0 are constants.  (A
    >detail: because ln(x) has an essential singularity at the origin, we 
    >can't set x0 to 0. Interesting question: what happens if x0-->0+?)
    
    	Let me take a stab at your question....
    
    	As X0--->0 then the numerator approaches zero and 0 / x is zero.
    
    	So the Wronskian appears to be zero at X0--->0+  ( As far as ln is
    	concerned, The lim X0--->0+  => ln (X0) ----> -oo )
    
    It's a shot in the dark,
    matt

     <<< Note 1512.4 by 3D::CORKUM "We'll be right there Miss Fletcher" >>>
                            -< A shot in the dark >-

    >which is of the desired form, since W(0) and x0 are constants.  (A
    >detail: because ln(x) has an essential singularity at the origin, we 
    >can't set x0 to 0. Interesting question: what happens if x0-->0+?)
    
<<    	Let me take a stab at your question....
    
<<    	As X0--->0 then the numerator approaches zero and 0 / x is zero.
    
<<    	So the Wronskian appears to be zero at X0--->0+  ( As far as ln is
<<    	concerned, The lim X0--->0+  => ln (X0) ----> -oo )

Note that we picked x0 to be an arbitrary real constant > 0.

Recall that W(x) = W(0) * x0 / x

If we let x0 tend towards zero, the Wronskian would tend toward zero for all x
(uniformly in any closed x-interval that does not contain zero); we're quite
sure (e.g., by blindly believing Abramowitz and Stegun) that the Wronskian is
*not* identically zero.

This seems to indicate that we should, in fact, have picked a specific value
for x0.  Exercise for the interested reader: which value should we have picked?

     <<< Note 1512.4 by 3D::CORKUM "We'll be right there Miss Fletcher" >>>
                            -< A shot in the dark >-

>>	(As far as ln is concerned, The lim X0--->0+  => ln (X0) ----> -oo)

No.  The logarithm has an essential singularity (i.e., a "pole of infinite
order") at the origin.  There is a theorem (by Picard, I believe) that says
that, given an analytic function f(z) with an essential singularity at z=z0,
f(z) assumes all values (except possibly one value) in any neighborhood of z0.

As another example, exp(1/z) has an essential singularity at the origin; it
assumes all values except zero in any neighborhood of the origin (regardless of
how small we make that neighborhood).